Management Aptitude Test 2015 Paper

(d ) Given, pipes A and B can fill a tank = 10 h,
B and C can fill it in 12 h,
A and C can fill it in 8 h.
Now, tank filled by all the pipes together
2(A + B + C ) = $\frac{1}{10}+\frac{1}{12}+\frac{1}{8}$
∵ LCM of 10, 12, 8 = 120.
= $\frac{1}{2}\left(\frac{1}{10}+\frac{1}{12}+\frac{1}{8}\right)$
= $\frac{1}{2}\left(\frac{12+10+15}{120}\right)$
= $\frac{1}{2}\times \frac{37}{120}$
∴ Pipe 8 alone can fill the tank in
= $\frac{1}{2}\times \frac{37}{120}-\frac{1}{8}$
= $\frac{370-300}{2400}$
= $\frac{70}{2400}=\frac{7}{240}$
Hence, B alone can fill the tank in $\frac{240}{7}$ h